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Need help analyzing a circuit

Discussion in 'Electronic Basics' started by James Howe, Dec 9, 2004.

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  1. James Howe

    James Howe Guest

    I'm new to electronics so I'm not great at circuit analysis. I've been
    reading some books and I understand some simple circuits, but the circuit
    below has me somewhat confused. This circuit is used to convert an input
    voltage into a constant current. The capacitor in the circuit is actually
    hooked into a timer to periodically discharge it, but I'm not interested
    in that part of the circuit at this time.

    Basically what I'm trying to figure out is voltage and current at parts of
    the circuit. I've tried my best to produce an ASCII version of the
    circuit, please excuse my crude attempt. The circuit basically consists
    of a handful of resistors, two 2N4126 PNP transistors, a .01uf capacitor
    and an LM324 op amp. The positive voltage is a control voltage and can
    range from approximately 0v to 5v. The negative voltage is fixed at -12v.

    In my reading, I've seen descriptions on how to compute voltage and
    current through NPN as well as PNP transitors. However, all the examples
    have a positive (or negative for PNP) voltage going into the collector and
    the emitter connected to ground. This circuit is different because the
    collector is connected to negative voltage through a capacitor and the
    emitter is connected to a positive voltage. I'm not sure the best way to
    attack this circuit for the purpose of figuring out voltage and current
    values. Any help or tips you can provide would be greatly appreciated.


    Here is the schematic:

    56k 1k
    +4v *--/\/\/\--+--/\/\/\----------------------+
    | |
    | / e
    | /
    | \ ------| Q1 (PNP)
    | -|\ /e \
    +------| \ / \ c
    | \--/\/\/\-|Q2 (PNP) |
    +------| / 47 \ |
    | +|/ \c |
    | / LM324 | |
    0v *----------+ +--------+
    |
    |
    |
    |
    .01 uf 100 |
    -12v *--------]|-------------------/\/\/\-------+


    And if the picture is munged, here is a description:

    A +4v supply is connected to a 56k and 1k resistor in series. Between the
    56k and 1k resistor a connection is made to the LM324 OPAMP on the
    inverting input. The non-inverting input is connected to ground. The
    output of the OPAMP connects to a 47 ohm resistor and into the base of a
    2N4196 PNP Transistor. The emitter is connected to another 2N4146 and the
    emitter of the second transistor connects to the other end of the 1k
    resistor. The collectors are tied together and are connected to a 100 ohm
    resistor which is connected to a .01 uf capacitor and finally to a -12v
    power supply. As I said earlier, in the full circuit, the capacitor is
    periodically discharged.

    Thanks.
     
  2. The main simplifying assumption to make is that the opamp will use its
    output to keep its two inputs at the same voltage. Since the
    noninverting input is at 0 volts, the output will use the output
    darlington pair of PNP transistors to hold its - (inverting ) input at
    zero volts, also.

    The second simplifying assumption is that the two inputs of the opamp
    draw no current.

    So any current that arrives at from the positive control voltage (+4)
    through the 56k resistor must be sucked up by the 1k emitter resistor
    in such a way that the node between those two resistors stays at zero
    volts. The two transistors are connected to multiply the current
    gains of the transistors, so that the base current of Q2 is so small
    that you can neglect it. So essentially all the current through the
    1k resistor also passes through the two collectors to the 100 ohm
    resistor. And this current equals the current through the 56k
    resistor with the control voltage minus zero volts across it,
    regardless of variations on the voltage on the collectors of the two
    PNP transistors. So the capacitor charges with a current that equals
    the current through the 56k resistor.
     
  3. A capacitor is a device that allows charge to 'pile up' (technical term).

    A pile of charge is a voltage.

    The formal relation is

    charge = capacitance times voltage

    This is another way of saying

    current = capacitance times 'rate of change of voltage'.

    For the above, you know that the opamp will try to keep the junction
    between the 56k and 1k resistor at 0V, so the current through the 56k
    resistor will be

    4V / 56k = 71.429uA

    The capacitor is 10nF, so the 'rate of change' of voltage is

    71.429uA / 10nF = 71.429e-6/10e-9 = 7.1429e3 volts/sec

    If the positive side of the capacitor starts at -12V, then to get to,
    say, -2V will take

    10/7.1429e3 = 1.4ms

    The waveform is constantly increasing (it's linear), so the output,
    taken from the positive side of the capacitor, is a 'ramp', which must
    be brought back to -12V periorically by shorting across the capacitor,
    probably using a transistor.

    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
  4. James Howe

    James Howe Guest

    Thanks for the explanation, but I have one question. I'm familiar with
    OPAMP's used with negative feedback and it sounds like this is what you
    are describing when you say that the OPAMP will attempt to keep the
    junction of th 56k and 1k at 0v. However, what confuses me is that the
    output of the OPAMP goes into two PNP transistors where the emitters are
    connected to the 1k resistor. In conventional electron flow, wouldn't the
    current be coming through the 1k, and through the emitters exiting through
    the base at the 47 ohm resistor and out through the collector connected to
    the 100 ohm resistor? In this arrangement I don't see how there can be
    negative feedback into the inverting pin since it would seem that the
    current couldn't flow through the base/emitter junctions back towards the
    1k resistor.

    I must be misunderstanding something.

    Thanks again.
     
  5. When the opamp output voltage is two diode drops more negative than
    the right end of the 1k resistor, the Q2 begins to conduct, supplying
    base current to Q1 which also begins to conduct, and all that current
    passes through the 1k resistor, making its right end more negative
    than its right. This process continues till the voltage at the
    junction of the two resistors balances at zero volts, so that all the
    current passing through the 56k resistor also passes through the 1k
    resistor, while the voltage across the 56k resistor is 4 volts.

    At the same time almost all that current also passes through the 100
    ohm resistor. The only part missing is the small base current being
    injected by the opamp output into Q2 which becomes part of the emitter
    current of Q1 but does not make it to the collectors.
     
  6. James Howe

    James Howe Guest

    Ok, that makes sense. Thanks for the explanation.
     
  7. One more thing. One nice way to get to play around with a circuit like
    this is to either build it and see what happens (not really possible
    unless you have a scope), or simulate it. Terry Pinnell has a great page
    on simulators (free and not free) that you can use. It's here

    http://dspace.dial.pipex.com/terrypin/ECADList.html

    I like circuitmaker student for simple stuff, and pspice student for
    temperature dependent analysis. There is a also a free one which people
    here tend to post with, called LTSpice, at the linear site.

    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
  8. James Howe

    James Howe Guest

    Funny you should mention building the circuit to just play around with as
    I was planning to do just that! In fact, I just picked up a used scope
    that I'm hoping to use for this sort of thing. I'll also look into the
    simulators. I had downloaded an evaluation copy of one and played around
    with it, but I like the idea of working with the real circuit.
     
  9. The problem with simulators is they tend to lie if you screw up the
    parameters, or if the circuit contains things that are difficult to
    model, or for any number of reasons. Also, you often get into situations
    where the mathematical models they use don't converge, and you end up
    trying to figure out ways to fix the circuit just to get it to converge.

    However, they let you 'see' things that are fairly difficult to see with
    a scope, like currents and power through circuit items. You can also
    show any number of varying voltages at once, thus overcoming the
    two-trace limitations of many oscilloscopes.

    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
  10. James Howe

    James Howe Guest

    Ok, I've got some really dumb followup questions. If the OPAMP is going
    to keep the junction between the 56k and 1k resistor at 0V, what is the
    purpose of the 47 ohm resistor and the Darlington pair? If the current
    through the 56k resistor basically is used to charge the capacitor, what
    role do the other components play? Why the 1k resistor (why that
    particular size)? Why the 100 ohm resistor?

    Thanks.
     
  11. I'm not sure I would have included those components. The 47 ohm
    resistor looks like a current limit resistor that comes into play if
    the collectors of the transistors have no current path (like if the
    capacitor is allowed to charge up till the first collector to base
    junction swings to forward bias). The 100 ohm resistor is more
    problematic, unless the discharge switch you mentioned earlier
    connects between it and the collectors, so that the peak discharge
    current is limited by the 100 ohms. You have to think about all the
    strange and unusual operating conditions the circuit might experience
    to guess what the designer had in mind for some components.
     
  12. James Howe

    James Howe Guest

    In the full circuit, there is a connection between the .01uf cap and the
    100 ohm resistor. The connection goes one way and connects to the
    non-inverting pin of another LM324 op amp, and the other direction
    connects to the discharge pin on a 555. The buffered output from the
    second op amp is fed through a 33k resistor and connected to the trigger
    and threshold pins on the 555. So the current goes through the 100 ohm
    resistor when the cap is charging, but the discharge does not. This
    particular circuit is used to create a voltage controlled oscillator.
     
  13. The darlington minimizes current flow into the opamp for control. With a
    normal PNP, you have a 1% error. With a darlington, you'll have a 0.01%
    error.

    You could minimize error even more using a p-channel JFET, which has
    essentially no leakage when the cap is charging, but also allows the
    current to be diverted when the cap reaches 0V. In order for this to
    matter, you need to use a good capacitor, such as a teflon cap.

    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
  14. In that case, I see no point to having the 100 ohm resistor. The
    transistors represent a limited source of current, whether charging
    the cap or dumping into the discharge pin of a 555.
     
  15. James Howe

    James Howe Guest

    That's interesting, could you explain that a little more? Is this
    something that is talked about in the data sheets for the 734 OPAMP? I
    read something somewhere last night about this op amp having problems at
    very low voltage levels. Does this configuration help the op amp produce
    accurate lower voltages. Is that what you meant by the 0.01% error?
    Also, would this configuration help with making the current converter more
    resistant to temperature issues? This particular circuit is part of an
    analog music synthesizer so they would want a particular voltage to
    produce a particular frequency with consistency.

    Thanks
     
  16. No, the issue I was referring to is current leakage for control. The way
    a bipolar transistor works is that some of the current is taken by the
    diode formed by the emitter-base (for PNP). That current flow allows a
    much larger current to flow from emitter to collector. For a normal
    transistor, the ratio of collector current to base current is called
    'beta'. It's usually between 15 and 300.

    Thus, if you are depending on the fact that the 56k resistor is
    delivering exactly 1/14000 A, some of that current will be diverted to
    the base rather than delivered to the capacitor. A ballpark figure is
    1/100, and this is where I got my 1% figure.

    For a darlington pair, however, the control current for the 'first'
    transistor is again diverted to the output, using 1/100 of *that*
    current. Thus, the real control current which flows into the opamp is
    1/100 of 1/100, or 1/10000 of the current that flows into the capacitor.
    Thus, the error required for control is 1/10000, or 0.01%.

    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
  17. Rich Grise

    Rich Grise Guest

    I once built an 8-input MUX for a scope, but this was in the days of 1 MHz
    clocks and stuff, i.e., I knew better than to use it for any kind of fast,
    or critical, stuff, but it did give a pretty good indication of what the
    circuit was doing! I think I was using it to watch the data bus on a
    micro I was debugging. Sort of a beer-budget logic analyzer. :)

    Cheers!
    Rich
     
  18. Rich Grise

    Rich Grise Guest

    I could see where it offloads some of the dissipation in the current
    source. Other than that, I have no clue.

    Thanks!
    Rich
     
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